Theorem sbc3or 39514

Description: sbcor 3677 with a 3-disjuncts. This proof is sbc3orgVD 39843 automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011.) (Revised by NM, 24-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbc3or	⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3or

Step	Hyp	Ref	Expression
1		sbcor 3677	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒))
2		df-3or 1109	. . . . 5 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
3	2	bicomi 216	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (𝜑 ∨ 𝜓 ∨ 𝜒))
4	3	sbcbii 3689	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒))
5		sbcor 3677	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))
6	5	orbi1i 938	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
7	1, 4, 6	3bitr3i 293	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8		df-3or 1109	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
9	7, 8	bitr4i 270	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 198   ∨ wo 874   ∨ w3o 1107  [wsbc 3633

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-3or 1109  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-v 3387  df-sbc 3634

This theorem is referenced by:  sbcoreleleq  39517